- #1

TehAdzMan

- 6

- 0

First post. I will attempt to use latex, something that involves me jabbing my keyboard with a pen since my \ key is missing.

We have an assignment question which I have solved, but there is a deeper issue I don't understand.

We are asked to find the complex Fourier series representation of

[tex] f(x) = x^2, \ -\pi < x < \pi, \ f(x+2\pi) = f(x) [/tex]

Initially I did the working, used integration by parts twice, and got the result

[tex] f(x) \sim \sum \sinh (in\pi)(\frac{\pi^2 n^2 + 2\pi in+2}{\pi i n^3})e^{-inx}) [/tex]

summed for n between negative infinity to infinity.

This is obviously incorrect because [itex] \sinh (in\pi) = 0 [/itex].

I received a tip off about how to go about solving the question, and early in the working I split the complex exponential term into its cos and isin terms.

Because the region of integration is symmetric and [itex] x^2 \times i\sin(nx) [/itex]

is odd, this term becomes 0 and the complex coefficients become real, the working proceeds and I get the correct answer which is

[tex]f(x) \sim \frac{\pi^2}{3} + 2\sum \frac{(-1)^n}{n^2} e^{-inx} [/tex]

where the sum is over all infinity again, except n = 0 which was worked out seperately.

I've omitted the working because it took so long just to do that, but if I need to I can show it.

I'm pretty sure the working in both cases is right.

What I don't understand is how they come to different answers, excluding for n = 0, which we work out separately, and is the same in either case.

The first way comes up with a sum of 0s, unless I've done something wrong.

I could understand if the first method returned an undefined answer.

Then it is just undefined and we have to try something else to get a defined answer.

But in both cases (excluding n = 0 obviously), the sums are well defined as far as I can tell.

Why are they different? Have I just done something wrong in the working or is there something I don't understand happening?

Adam.